Frequency Source Project

[donniegunz]

Hey Guys,
I have been putting off this project for months but want to get it done by the end of the year. I was wondering if anyone has any ideas how to convert a TTL signal to AC (aka crossing zero)? I am using one of the those cheap square wave oscillator kits you can find on any hobby website. The key to this project is to be able to choose between AC or TTL. I need full functionality of both. I know there has to be a relatively simple solution to this but it I don't have the aptitude to figure it out. I have attached the schematic of the TTL source I am using. Any help is appreciated!

 

[jpanhalt]

How much current do you need?

John

 

[donniegunz]

Less than 1A works for me but 20ma is ideal.

 

[4pyros]

Does the AC need to go below 0 volts?

 

[panic mode]

just connect capacitor to output (right of LED)....

 

[donniegunz]

@4pyros - Yes it needs to cross zero.

@panic mode - Simply connecting a capacitor will make a TTL signal become AC?

 

[panic mode]

in general we can consider ANY signal as composition of AC (shape) and DC (vertical shift).
running any signal through capacitor, removes DC component and all you get is AC.

 

[donniegunz]

@panic mode - I understand what you are saying however what I need is zero-crossing AC. 10v p-p for example +5v and -5v. True AC. If what you are saying is accurate there would be no need for inverters. I appreciate your help. Should I be more detailed with what I am trying to accomplish?

 

[donniegunz]

Thanks for all the help guys. I found my answer on a previous thread.

https://www.electro-tech-online.com/threads/square-wave-with-zero-crossing.92136/

ericgibbs is an intellectual badass indeed!

 

[crutschow]

in general we can consider ANY signal as composition of AC (shape) and DC (vertical shift).
running any signal through capacitor, removes DC component and all you get is AC.

It's important to note that you will get a symmetrical plus and minus signal from capacitor coupling only if the duty-cycle of the signal is 50%.

 

[panic mode]

true, but there was not much mention of kind of AC output.
using opamp and dual power supply can solve the problem.
if Vref is replaced with potentiometer (0..5V) then one can freely change
DC bias of output AC.
opamp must be rail to rail (both inputs and outputs) or supply need to be higher.

 

[Nigel Goodwin]

true, but there was not much mention of kind of AC output.
using opamp and dual power supply can solve the problem.
if Vref is replaced with potentiometer (0..5V) then one can freely change
DC bias of output AC.
opamp must be rail to rail (both inputs and outputs) or supply need to be higher.

Bit of overkill, when a simple capacitor can do the job

I would imagine a bigger 'problem' is the fact it's a squarewave, so there's not so much of a 'zero-crossing' point to try and detect, it's simply whenever the logic level changes.

 

[alec_t]

If the fact that it's a square-wave is a problem then apply the square-wave to an RC integrator to get a triangle-wave and apply the triangle-wave via a high-value resistor and series capacitor to a pair of back-to-back diodes. The voltage across the diodes will be a pseudo-sinewave with recognisable zero-crossing points.
Check out https://www.electro-tech-online.com/threads/taillight-sequencer.492/

 

[panic mode]

@donniegunz

no problem. i didn't know what is your experience of comfort level, i read 'simple' in the first post so i offered the simplest possible solution. using opamp has other advantages, like ability to drive smaller impedance loads etc.



as crutschow noted, your oscillator does not produce perfectly symmetrical signal so he is correct that AC after capacitor would not have absolutely equal amplitudes for positive and negative part of period.

the slight assymetry is due fact that in this oscillator capacitor is charged through R1+R2=71.9k, but discharged only through R2 (68k).
so you get square wave with duty cycle that is 48.6% because
68/(68+71.9)=0.4860617... (on time is 48.6%, off time is 51.4%).

this 1.4% in duty corresponds to some 70mV offset. (integrals or areas of positive and negative halfperiod are equal, so if one pulse is wider, it will be lower in amplitude and the other way arround).

i seriously doubt that your application would suffer because you get +5.07 to -4.93V insted of +5V...-5V.

even this can be adjusted by adding biasing circuit of two resistors or simply one potentiometer (in which case you can adjust DC offset to your liking). for example one side of potentiometer could connect to +5V, othr to -5V and wiper to output (after coupling capacitor).